Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $p:Y\to X$ be a birational proper surjective morphism of regular surfaces, and let $D$ be a divisor on $Y$ such that $p(D)$ is a point. Then $p_\ast D =0$ by definition. Is there an easy way to show that $p_\ast \mathcal{O}_Y(D) = \mathcal{O}_X$?

More generally, does the equality $c_1(p_\ast L) = p_\ast c_1(L)$ hold, where $L$ is a line bundle on $L$? Or equivalently, does the equality $\mathcal{O}_X(p_\ast D) = p_\ast \mathcal{O}_Y(D)$ hold, where $D$ is a divisor on $Y$?

I just found out the answer is positive, but my proof is a bit complicated I think. So the question remains, but now I'm looking for an easy proof.

share|improve this question
    
If $D<0$ and is mapped to one point of $X$, then $p_*O_Y(D)=0$. –  user18119 Mar 13 '13 at 23:45

1 Answer 1

up vote 2 down vote accepted

Here is an idea:

The proper birational morphism $p$ can be factored as a finite sequence of blow-downs of $(-1)$-curves, according to Beauville's Complex Algebraic Surfaces Theorem II.11. Thus, saying that $p(D)$ is a point is equivalent to saying that $D$ is a union of exceptional divisors of the blow-downs. By Proposition II.3 of the same book, we know that $\operatorname{Pic}(Y) = \operatorname{Pic}(X)\oplus\mathbb Z^n$ where $n$ is the number of blow-downs of a single $(-1)$-curve (and the obvious generators of the $\mathbb Z^n$ part are the $(-1)$-curves blown down at each step). Since $D$ is a union of exceptional divisors, we know that $\mathcal O_Y(D) = (0,a_1,\ldots,a_n)\in\operatorname{Pic}(Y),$ i.e., the $\operatorname{Pic}(X)$ component of $D$ in $\operatorname{Pic}(Y)$ must be zero. This tells us that $p_*\mathcal O_Y(D)=\mathcal O_X.$

I believe the same method will work for the more general case, using the definition of $p_*D$ and the fact that $p_*\mathcal O_Y(D)$ will be just the $\operatorname{Pic}(X)$ component of $\mathcal O_Y(D).$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.